
A toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$. How many revolutions will the smaller wheel make when the larger one makes $15$ revolutions?
A. $25{\text{ revolutions}}$
B. $20{\text{ revolutions}}$
C. ${\text{1 revolution}}$
D. $10{\text{ revolutions}}$
Answer
523.5k+ views
Hint The distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ .
The distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different.
Complete step by step answer
As given in the question that the toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$ .
Let the number of revolutions smaller wheel will make when the larger one makes $15$ revolutions be $n$
We know that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$
So, the distance covered by larger and smaller wheels in one revolution will be $50\pi $ and $30\pi $ respectively.
Now, the distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different. So,
${\text{Distance covered by larger wheel in 15 revolutions }} = {\text{ Distance covered by smaller wheel in n revolutions}}$After substituting the values we have
$15 \times 50\pi = n \times 30\pi $
On simplifying we have
$n = \dfrac{{15 \times 50}}{{30}} = 25$
Therefore, the smaller wheel will make 25 revolutions when the larger one makes 15 revolutions.
Hence, option A is correct.
Note We have used the fact that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ . Although it is applicable only when the wheel is in uniform circular motion and there is no angular acceleration. In uniform circular motion, the magnitude of velocity of each particle in motion is constant throughout the motion.
Angular acceleration is basically the time rate of change of angular velocity and the angular velocity is the rate of change of angular displacement.
The distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different.
Complete step by step answer
As given in the question that the toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$ .
Let the number of revolutions smaller wheel will make when the larger one makes $15$ revolutions be $n$
We know that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$
So, the distance covered by larger and smaller wheels in one revolution will be $50\pi $ and $30\pi $ respectively.
Now, the distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different. So,
${\text{Distance covered by larger wheel in 15 revolutions }} = {\text{ Distance covered by smaller wheel in n revolutions}}$After substituting the values we have
$15 \times 50\pi = n \times 30\pi $
On simplifying we have
$n = \dfrac{{15 \times 50}}{{30}} = 25$
Therefore, the smaller wheel will make 25 revolutions when the larger one makes 15 revolutions.
Hence, option A is correct.
Note We have used the fact that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ . Although it is applicable only when the wheel is in uniform circular motion and there is no angular acceleration. In uniform circular motion, the magnitude of velocity of each particle in motion is constant throughout the motion.
Angular acceleration is basically the time rate of change of angular velocity and the angular velocity is the rate of change of angular displacement.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

